Support the Arctic Sea Ice Forum and Blog ### Author Topic: Applicability of Stefan-Boltzmann Law to planets as a whole  (Read 6590 times)

#### F.Tnioli

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« on: August 18, 2014, 03:24:08 PM »
I am no specialist in astrophysics, and i am seeking some basic understanding about whether (and how exactly) Stefan-Boltzmann Law could be applicable to rocky planets with a significantly thick athmosphere.

Few days ago, i've found a link (provided by someone in this forum; thanks!) to a simple calculator: http://www.endmemo.com/physics/radenergy.php . Entering values into the thing confirms one of my old amateurish conclusions: that considering amount of solar radiation reaching Earth and possible albedo and greenhouse effect changes, it is very unlikely to see some +10C warming (for the average annual surface temperature of Earth), and quite impossible to see some +20C or higher (both values are above-pre-industrial, of course).

For said conclusion to be valid, though, one has to assume that Earth's athmosphere is quite transparent - in both directions. Yet, we know that both water vapour and CO2 allow most of Sun's energy in (visible spectrum), but does not (as easily) allow most of Earth's own emission out - since it's in IR spectrum.

And then i had this idea: well, we have a practical example of an extremum of greenhouse athmpshere: Venus. 96,5% of its athmosphere is CO2 (with the rest being 3,5% nitrogen and some trace amounts of other gases), and it's darn thick and heavy athmosphere with that - some 9+ MPa pressure at the surface (Earth has ~0,1MPa surface pressure). What if i'd try to use the calculator to see how much different from Stefan-Boltzmann Law temperature such an athmosphere would produce at the planet's surface? Assuming that long-term, incoming energy = emitted energy in terms of the whole planet, of course.

Here are assumptions and numbers i've got when i tried:

Venus has (known from measurements):
(Bond) Albedo = 0,9  // which is drastically different from Earth's 0,306
ergo, emissivity = 0,1 (Kirchhoff's law)
surface area = 4,6E+14 meters
average surface temperature = 737 K

Using the calculator, we get: 7,7E+17 W //that's her "would be output" roughly

We also know that for Venus:
solar constant = 2611 W/m^2
crosssection = 1,15E+14 m^2 //not counting extra size her athmosphere is
ergo, input = 2611*1,15E+14*(1-0,9) = 3,004E+17 W

Obviously, Venus can not (long-term, at least) emit more than twice amount of energy than it absorbs. Any output from its core - i mean fission of uranium-like radioactive matherials, - can't be of this order of magnitude after some 4+ billions years of the planet's existance in solid form. Therefore, the only possible explanation - is that temperature of Venus' hard surface is higher because of lack of transparency of its athmosphere. If by any chance Venus would have no athmosphere whatsoever, yet its surface would still have the same 0,9 albedo (all wavelengths average) the planet has today with its athmosphere present, - then its surface temperature would be lower. How much lower? sqrt(sqrt(7,7/3,0)) lower, since it's 4th power of the temperature in the formula; i.e. 1,2657 times lower: 737 / 1,2657 = 582,3 K. Therefore, Venus' athmosphere greenhouse effect warms up the surface by 737 - 582,3 = 154,7 degrees Kelvin. No wonder, with so much higher than Earth's solar constant (Earth has it 1366,1W/m^2 iirc), and with such a "greenhouse hell" athmosphere.

The difference is significant, but not awfully dramatic. Back to Earth, our athmosphere has some ~90 times lower pressure at the planet's surface than Venus', and CO2 % content is ~250 times lower. I'd say, Stefan-Boltzmann Law is much more applicable to Earth directly, with greenhouse effect giving relatively minor effect.

Same calculations as above, but this time for Earth, albedo 0,306, surface temperature 288 K, surface area 5,1E+14 m, solar constant 1366,1 W/m^2, crosssection 1,275E+14 m^2:

"would be output" for Earth = 1,38E+17 W
actual input for Earth = 1,21E+17 W

Once again, the crosssection did not include the extra radius given by the athmosphere; this explains some of the difference. The rest (and the most) is the greenhouse effect.

Which is, finally, where i get to a question. Problem is, "official" strength of the greenhouse effect on Earth - is some ~33 K warmer than otherwise expected. This is supposed to be good science. According to the calculator, "would be output" of 33 K colder Earth (so it's 255 K for the temperature, emissivity = 1 - albedo = 0,694) - would be 8,5E+16 W, which is some 30% LESS than actual input for Earth calculated just above. Which seems quite impossible (for a body without an athmosphere and with a given bond albedo / emissivity). If to estimate Earth's greenhouse effect using the numbers above, then it seems the greenhouse effect is just 8 to 9 degrees kelvin - some ~4 times less than the official amount of +33 K. So the question is:

where's an error in my amateurish attempts to apply Stefan-Boltzman to Earth? What am i missing? ><

Thank you in advance for any help...

« Last Edit: August 18, 2014, 03:32:24 PM by F.Tnioli »
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#### crandles

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« Reply #1 on: August 18, 2014, 04:02:15 PM »
Not sure I understand question.

Did some calculations on an MIT course.

Solar constant 1370 W/m^2
SB constant 5.67E-8 W/m^2/K^4
Albedo earth 0.3
Ratio of surface area of sphere to that of circle with same radius (that of earth) =4

(Your 1366.1 and .306 are likely more accurate. Easier to use factor 4 than surface area 5,1E+14 m and crosssection 1,275E+14 m^2)

Anyway
1370/4*(1-0.3)=239.75 W/m^2

239.75= SB* T^4

239.75/5.67E-8=4.228E+9

4.228E+9^0.25 = 255k effective emission temperature

Average temperature of Earth ~15C =273+15=288k

actual temperature 288k - effective temp with no atmosphere 255k = 33k Greenhouse effect

I don't see where you are getting 1.38E+17 values from.
« Last Edit: August 18, 2014, 04:21:01 PM by crandles »

#### Wipneus ##### Re: Applicability of Stefan-Boltzmann Law to planets as a whole
« Reply #2 on: August 18, 2014, 04:07:26 PM »

Here are assumptions and numbers i've got when i tried:

Venus has (known from measurements):
(Bond) Albedo = 0,9  // which is drastically different from Earth's 0,306
ergo, emissivity = 0,1 (Kirchhoff's law)

At least this is wrong, Kirchhoff's law is only strictly true for a single frequency/wavelength.

Assuming emissivity = 1 is probably a better assumption, although I could not really be sure about Venus clouds.

#### F.Tnioli

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« Reply #3 on: August 18, 2014, 04:12:17 PM »
...
I don't see where you are getting E+16 or E+17 values from.
No time to work on your numbers right now, but E+16 and E+17 are values from the calculator - the link is given, - when using said values. For example, if you'll use (in the calculator) 255 K for temperature, .7 for emissivity (which is 1 - albedo_of_Earth), and 510000000000000 meters for surface - which is 5,1E+14 if to put it in this form - then the calculator will give you a number with E+16 as an "answer" in Watts. Etc.

Laters.
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#### crandles

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« Reply #4 on: August 18, 2014, 04:27:43 PM »
...
I don't see where you are getting E+16 or E+17 values from.
No time to work on your numbers right now, but E+16 and E+17 are values from the calculator - the link is given, - when using said values. For example, if you'll use (in the calculator) 255 K for temperature, .7 for emissivity (which is 1 - albedo_of_Earth), and 510000000000000 meters for surface - which is 5,1E+14 if to put it in this form - then the calculator will give you a number with E+16 as an "answer" in Watts. Etc.

Laters.

Sorry figured out 239.75 W/m^2 * surface area 5.1E+14 m^2 = 1.22E+17 W

Not clear on where 1.38E+17 comes from. (Seems to be about 1.22E+17 *288/255)

#### F.Tnioli

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« Reply #5 on: August 19, 2014, 08:52:08 AM »

Here are assumptions and numbers i've got when i tried:

Venus has (known from measurements):
(Bond) Albedo = 0,9  // which is drastically different from Earth's 0,306
ergo, emissivity = 0,1 (Kirchhoff's law)

At least this is wrong, Kirchhoff's law is only strictly true for a single frequency/wavelength.

Assuming emissivity = 1 is probably a better assumption, although I could not really be sure about Venus clouds.
Thanks for explanation about Kirchhoff's, i suspected it'd be somewhat different, myself; but not making the simplyfing assumption to equate the two - would mean the calculation with any hope of even some little precision would become impossible (for me), because i am unable to define the difference; so i had to make the assumption and to hope the difference between the two won't be more than a few percents. Thus i still doubt this could be "the" answer to my question, since in case of Earth, i still doubt that the difference between "absorbivity" (to calculate actual input) and emissivity (to calculate "would be output" using the calculator i mentioned) could be as large as some ~30% mentioned in the question (S-B law has emissivity in the power of 1, so emissivity is directly proportional in terms of how much it affects energy output calculated). Or could it be that large?..

Sorry figured out 239.75 W/m^2 * surface area 5.1E+14 m^2 = 1.22E+17 W

Not clear on where 1.38E+17 comes from. (Seems to be about 1.22E+17 *288/255)
The last line is probably correct, but it's "backwards" way to get it if to speak about the 1st post. 1,38E+17 originaly calculated via - quote - "same calculations as above" while using numbers for Earth (instead of Venus). I.e., the number is also a result from the calculator i gave the link to, when using the following numbers to calculate "would be output" (from S-B law): emissivity = 0,694 (assuming - as Wipneus pointed out, very unreliably - that emissivity equals ( 1 - albedo ) for Earth, i.e. 1 - 0,306 = 0,694); surface area = 5,1E+1 (i.e. 510000000000000 meters); surface temperature = 288 K (i.e. +15C). I just entered these values into the calculator once again to verify if there is any error in the number i've wrote down in the 1st post, and it still appears i didn't typo or something - once again, i get 1,38E+17 (or if to report it with all the digits calculator produces - it's 1.3806470092869E+17).

As for the difference between yours 1,22E+17 and 1,21e+17 in the 1st post - (edited) - probably comes from a bit different numbers you've used in your 1st post in the topic.

To answer my initial question, some extra good-science data about actual Earth energy output would be nice. The figure on the page 2 of this paper ( http://www.aos.wisc.edu/~tristan/publications/2012_EBupdate_stephens_ngeo1580.pdf ) seems to be helpful for that. It says Earth radiates out 239,7 W/m^2 (some +-1,4% uncertainty). I guess this is 24/365 average for the whole planet. Then total energy radiated would be 239,7 W/m^2 * 5,1E+14 m^2 = 1,22247E+17 W. Which is still some ~30% higher than amount of energy which would be radiated out by Earth with 0,694 emissivity at 255K temperature and without any athmosphere, if to trust Stefan-Boltzmann law (which is what Earth would be without the greenhouse effect, they say, and assuming its emissivity is not some ~30% higher than 0,694). Is the latter assumption the problem indeed? Would Earth-without-athmosphere emissivity be actually some ~0,9? >< I still doubt that; some 71% of Earth surface is water, and on the calculator's page, water is said to have 0,67 emissivity. So it gotta be something else. What? ><
« Last Edit: August 19, 2014, 12:31:08 PM by F.Tnioli »
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#### crandles

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« Reply #6 on: August 19, 2014, 01:45:04 PM »

The last line is probably correct, but it's "backwards" way to get it if to speak about the 1st post. 1,38E+17 originaly calculated via - quote - "same calculations as above" while using numbers for Earth (instead of Venus). I.e., the number is also a result from the calculator i gave the link to, when using the following numbers to calculate "would be output" (from S-B law): emissivity = 0,694 (assuming - as Wipneus pointed out, very unreliably - that emissivity equals ( 1 - albedo ) for Earth, i.e. 1 - 0,306 = 0,694); surface area = 5,1E+1 (i.e. 510000000000000 meters); surface temperature = 288 K (i.e. +15C). I just entered these values into the calculator once again to verify if there is any error in the number i've wrote down in the 1st post, and it still appears i didn't typo or something - once again, i get 1,38E+17 (or if to report it with all the digits calculator produces - it's 1.3806470092869E+17).

You shouldn't be using 288k in this calculation to arrive at 1.38E+17.

There is just one formula and therefore just one route to the effective emission temperature of 255k.

Solar constant*(1-albedo)/ratio of surface area of sphere to circle= SB * T^4

Using available information you can solve the equation to work out T=255K

So if the temperature is 255K then the inputs and output match. Thus you should use 255k not 288k to work out the output Watts for a black body earth (i.e. with completely transparent atmosphere).

If you insist on using area of earth to get to Watts than than keeping it simple and just using W/m^2 and ratio of surface area of sphere to circle:

Then SB * T^4 * surface area of Earth = output watts
5.67E-8*255^4*5.1E+14 = 1.22E+17

also solar constant * (1-albedo) * area of circle with Earth radius = input watts
1370*0.7*1.275E+14 = 1.22E+17

These match so the 255K is correctly calculated and given the current average temperature is 288k there is a 33C greenhouse effect.

(I am still not really sure how you are using 288k to get to 1.38E+17 because SB*T^4*area of earth=1.99E+17 Watts if you use a temperature of 288k instead of using the correct 255k. If you are using (1-albedo) then you appear to be calculating input watts and 288k shouldn't play a part. So it looks like a misguided calculation and I still can't work out what you are trying to calculate and how in order to try to explain the error. That is a minor issue if you can see that above shows that 255k is correctly calculated so that input and output watts match.)
« Last Edit: August 19, 2014, 02:40:37 PM by crandles »

#### F.Tnioli

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« Reply #7 on: August 20, 2014, 09:49:24 AM »
You shouldn't be using 288k in this calculation to arrive at 1.38E+17.
...
I am not using 288k "to arrive at 1.38E+17". I am using it for something else, and i believe - correctly. Namely, i am using 288k to estimate how applicable Stefan-Boltzmann law is to a planet with a thick athmosphere. This is the essense of this topic. You see, we know average actual Earth surface temperature = 288k (from measurements - satellites records, etc). Based on that, i calculate how much energy such a surface - at such a temperature, - would be radiating out according to Stefan-Boltzmann law (and using the calculator via the link provided, to simplify the process).

Then i calculate actual input (using Earth crosssection and solar constant) and compare the two figures. In case of positive greenhouse effect, the former figure will be higher than the latter. How _much_ higher - is exactly what i was interested to find; i compared such a difference for Venus and for Earth, and numbers i've found are imho quite interesting to see. In case of anti-greenhouse effect (i.e. negative change of surface temperature due to athmospheric processes - as observed on few other bodies in the Solar system), the former figure would be lower than the latter.

This is what 1st post is about before it gets to the question.

... There is just one formula and therefore just one route to the effective emission temperature of 255k.
I disagree. There are often more than one way to calculate something in physics, and this case - i think - is no exception. On top of the way you describe, there is also this path: 1. measure/calculate actual input of energy for "Earth with completely transparent athmosphere" - how much Earth would absorb from the sun; this should be quite close to the amount Earth absorbs in reality, since actual Earth athmosphere is quite transparent to visible spectrum, and most of Sun's energy comes in in exactly that - quite short-wave - form; 2. knowing that for any long-term situation, amount of absorbed energy _have to_ equal amount of radiated energy (geothermal is a microscopic share, and can be discarded for simplicity, as can be background space radiation, energy from meteorite impacts, etc), - use Stefan-Boltzman law "in reverse": calculate what temperature Earth surface (without any athmosphere) will be while knowing what energy it will be radiating out (which, again, has to be equal with how much it'd absorb, long-term).

I've done it even when writing 1st post, and as a result, i've found that Earth greenhouse effect should be - quote - "it seems that greenhouse effect is some 8...9 degrees". Namely, if you'd put T=279.3 K, surface area of Earth and emissivity = 0.694 into said calculator and hit "calculate", then the result you get - is 1.2212271426758E+17 . Which is a match to the amount of energy Earth absorbs. The difference between 279.3 K and 288 K is exactly "some 8...9 degrees kelvin" i've talked about; some ~4 times less than the "official" 33 K for greenhouse effect. See?

... output Watts for a black body earth (i.e. with completely transparent atmosphere). ...
Never was calculating anything for a "black body" earth; this would mean emissivity ~= 1, albedo ~= 0. As mentioned above, actual Earth is some ~70% water at its surface, and water has emissivity of some ~0.67. Definitely not a black body to me.

... Then SB * T^4 * surface area of Earth = output watts
5.67E-8*255^4*5.1E+14 = 1.22E+17 ...
This is correct for a black body. But not for reality of Earth. Since Earth is not a sphere covered by some thick layer of soot everywhere on its surface. Water is 0,67 emissivity, sand (Earth has vast deserts) is 0.76 emissivity, areas covered with thick layers of green plants - i don't know, but far lower than 1.0 i bet, soil surface - 0.38, clay - 0.39. I've used 0.694 in the 1st post as average emissivity for Earth surface. And got proportionally lower value as a result. From which my original question appears, of course.

... also solar constant * (1-albedo) * area of circle with Earth radius = input watts
1370*0.7*1.275E+14 = 1.22E+17 ...
This is exactly what i've done in the 1st post for both Venus and then Earth. Slightly different value for solar constant i've used (1366.1) and for albedo (0.306) are the only (cosmetic) differencies, due to which the result found in the 1st post is 1.21 (not 1.22) E+17 W. "Crosssection" and "area of circle with Earth radius" - is the same thing; crosssections for both Venus and Earth in the 1st post were calculated via "radius squared multiplied by Pi", with Pi=3.14159265 (yes i do remember it with this precision, if you wonder). Numbers for Venus and Earth crosssections in the 1st post are exactly the product of this very basic calculation.

... These match so the 255K is correctly calculated and given the current average temperature is 288k there is a 33C greenhouse effect. ...
So you calculated how much energy "black body" Earth would radiate out, defined a temperature at which the amount of radiated (by the black body Earth) energy would equal amount of energy absorbed by a NOT black-body Earth energy (since you used 1-albedo in the SECOND part, but not in 1st), - and based on that, you define "greenhouse effect". Right? =)

Excuse me, but how exactly removal of the athmosphere automatically produces a decrease of Earth albedo from the value you used to ZERO? =) And, of course, the Kirchhoff's law, while being strictly true only for an isolated frequency, - still plays a major role even if we speak about all frequences; this means, if you talk about black body emitting energy, then you _have to_ automatically assume that such a body's albedo is ZERO, and use this value when you calculate absorbed by such a body energy, as well.

Alternatively, if you use actual albedo to calculate how much energy a body actually absorbs - then you _have to_ add corresponding emissivity when you calculate how much energy such a body would radiate out. And to do it quantatively right (with precision) - is difficult, as Wipneus pointed out, since bodies often radiate in a different spectrum than spectrum they absorb at (making calculating exact effect of Kirchhoff's law a huge mess!).

... If you are using (1-albedo) then you appear to be calculating input watts. ...
Nope - not nesessarily. Again, Kirchhoff's law; plus an assumption i used - which was, that said law is directly and fully applicable to a planet as a whole (which is possibly rather big error to my calculations, but not on the order of dozens percents, i dare hope) - allowed me to find _emissivity_ of a body as (1 - albedo). The word _emissivity_ itself indicates that such a calculation is not about "input watts", but about "output watts".

... So it looks like a misguided calculation and I still can't work out what you are trying to calculate and how in order to try to explain the error ...
Hopefully, this post will allow you to see that it wasn't misguided calculation, and will allow you to work out what i was calculating.

P.S. when i am saying "Kirchhoff's law", i, naturally, mean this one: http://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation . In particular, this corollary from it (my bold): ...  the emissive power of an arbitrary opaque body of fixed size and shape at a definite temperature can be described by a dimensionless ratio, sometimes called the emissivity, the ratio of the emissive power of the body to the emissive power of a black body of the same size and shape at the same fixed temperature. With this definition, a corollary of Kirchhoff's law is that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity. ... and, of course, absorptivity is nothing else than (1 - albedo), by the albedo's definition.
« Last Edit: August 20, 2014, 12:40:51 PM by F.Tnioli »
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#### F.Tnioli

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« Reply #8 on: August 20, 2014, 12:56:16 PM »
Oh and by the way, one of quite nice consequences of numbers calculated in the 1st post - is this:

since temperature parameter in Stefan-Boltzmann law is in power of 4, using my numbers, we can find the error which is produced when we simply _discard_ greenhouse effect of the Earth's athmosphere and apply S-B law (in reverse) to Earth directly, in order to calculate its surface temperature for any given energy input. This error will be:

(T)error = ( sqrt ( sqrt ( 1,38E+17 W / 1,21E+17 W ) ) - 1 ) * 100% = 3,34% . Which is amazingly low. This means, applying S-B law to Earth but "forgetting" to account for greenhouse effect when doing so - produces wrong _temperature_ (for a given energy input), but such a wrong result is only ~3,34% away from reality (for Earth, of course - Venus is very different in this regard). Still, for Earth - nice, isn't it.

And yes, this has practical use: with solar input practically constant in terms of some decades, but with growing imbalance caused by growing greenhouse effect, "net input" changes; applying reversed S-B law for such actually increasing net-input energy, we can roughly estimate how much surface temperature will increase. With such relatively low error (some ~3,5%), this simple method to do so is good for at least comparing numbers produced by it with any numbers produced by complex models and such. And, of course, this error fully applies if we use S-B law to actually "convert" either energy absrobtion or emission into temperature, or vice versa - temprerature into energy absorbtion or emission. But, if we use S-B law to actually compare any two given surface temperatures in terms how much more energy (in percents) Earth would need to absorb/radiate to reach the higher one given temperature, - then most of this error won't be present (since when calculating percentages, this error, however large, will be present both above and below main " / " (divider) of the equation, with only very minor difference as long as two given temperatures are quite close (such as, for example, 288K and 298K), and thus, most of error will self-destroy; like, having 3,34% above the divider and some 3,36% below the divider - results in remaining (small) error of (3,34%/(3,34%*1,006)) = 1/1,006 = 0,994, i.e. 0,6%. Even nicer, isn't it. « Last Edit: August 20, 2014, 01:19:52 PM by F.Tnioli »
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#### crandles

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« Reply #9 on: August 20, 2014, 02:13:31 PM »
use Stefan-Boltzman law "in reverse": calculate what temperature Earth surface (without any athmosphere) will be while knowing what energy it will be radiating out (which, again, has to be equal with how much it'd absorb, long-term).

I've done it even when writing 1st post, and as a result, i've found that Earth greenhouse effect should be - quote - "it seems that greenhouse effect is some 8...9 degrees". Namely, if you'd put T=279.3 K, surface area of Earth and emissivity = 0.694 into said calculator and hit "calculate", then the result you get - is 1.2212271426758E+17 . Which is a match to the amount of energy Earth absorbs. The difference between 279.3 K and 288 K is exactly "some 8...9 degrees kelvin" i've talked about; some ~4 times less than the "official" 33 K for greenhouse effect. See?

That 0.694 is the wrong value.

https://en.wikipedia.org/wiki/Effective_temperature
Quote
Let's look at the Earth. The Earth has an albedo of about 0.367. The emissivity is dependent on the type of surface and many climate models set the value of the Earth's emissivity to 1. However, a more realistic value is 0.96.

Ref 9 being
http://terpconnect.umd.edu/~sliang/papers/Jin2006.emissivity.pdf

(The MIT course I did set it to 1.)

#### crandles

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« Reply #10 on: August 20, 2014, 02:29:33 PM »
http://modis.gsfc.nasa.gov/sci_team/meetings/200503/posters/ocean/minnett1.pdf

provides some measurements for ocean emissivity.

It seems the wavelength dramatically changes the albedo & emissivity. For 6000k temperature solar radiation albedo ~0.3 but with Earth's outgoing radiation, circa 15C, emissivity is well over 0.9.

Sorry I didn't answer you well earlier.

#### viddaloo

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« Reply #11 on: August 21, 2014, 01:00:58 AM »
If to estimate Earth's greenhouse effect using the numbers above, then it seems the greenhouse effect is just 8 to 9 degrees kelvin - some ~4 times less than the official amount of +33 K. So the question is:

where's an error in my amateurish attempts to apply Stefan-Boltzman to Earth? What am i missing?

As always it's way too easy to get lost in the numbers and the details... Been there, done that, haha! I'd say — honestly and in ernest — you may be missing the 4th dimension here. Ie Time. Let's say you — like me — first heard about the Greenhouse Effect 1986–ish. At that time GE wasn't so big, it wasn't such a big deal. If it had stayed like that for centuries and eons, only die–hard nerds would even care about the effect. (And besides, natural and pre–industrial GE is after all part of what makes life liveable, at least here in the chilly Nordic countries.)

Add the 4th dimension, and you get a *timeline* of events. Perhaps with the Arctic going ice–free in the summer of 2015–2020 for the very first time. In turn setting off further warming and feedback mechanisms. Or if you're a math genius, then think of an exponential curve, rising extremely slowly for the first 50 or hundred years. This is the sort of timeline we're on today, IMO. A line that bends. More and more. As it traps more and more heat. Like Venus.
« Last Edit: August 21, 2014, 08:12:34 PM by viddaloo »
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#### F.Tnioli

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« Reply #12 on: August 21, 2014, 09:25:18 AM »
http://modis.gsfc.nasa.gov/sci_team/meetings/200503/posters/ocean/minnett1.pdf

provides some measurements for ocean emissivity.

It seems the wavelength dramatically changes the albedo & emissivity. For 6000k temperature solar radiation albedo ~0.3 but with Earth's outgoing radiation, circa 15C, emissivity is well over 0.9.

Sorry I didn't answer you well earlier.
Quite very interesting. The paper has a table on its 3rd page "for referense", which among other things says that emissivity of water is .95. The calculator i gave the link to in the 1st post - lists water as .67 emissivity. Now, how about some other place listing it as -15, and some fourth insisting its actually equals "drunk purple butterfly", whatever it'd mean... Sigh.

I take it that whomever was typing emissivity values listed within that calculator was using numpad to type in numbers... Since on the numpad, button "6" is next to "9", so then 0.67 instead of 0.97 could be an easy typo to do, eh.

At least i got an answer to my question, though.

Thank you.
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#### F.Tnioli

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« Reply #13 on: August 22, 2014, 12:03:27 PM »
Continuing to educate myself on the subject and beyond, i found one extremely helpful and informative piece, which brings in lots of good "few napkins" science on the subject, as well as lots of relevant data. Granted, there are approximations and simplifications where reasonable (in particular, Earth emissivity is assumed to be 1, not some 0,96 or so). However, in general, it is so good i decided to make this separate post just for it. Especially useful are bits about optical thickness.

Without further ado - http://bartonpaullevenson.com/NewPlanetTemps.html .
To everyone: before posting in a melting season topic, please be sure to know contents of this moderator's post: https://forum.arctic-sea-ice.net/index.php/topic,3017.msg261893.html#msg261893 . Thanks!